v2009.01.01 - Convex Optimization

A.7. ZEROS 571
(TRANSPOSE.)
Likewise, for any matrix A∈ R m×n
rank(A T ) + dim N(A T ) = m (1471)
For any square A∈ R m×m , the number of 0 eigenvalues is at least equal
to dim N(A T ) = dim N(A) while the left-eigenvectors (eigenvectors of A T )
corresponding to those 0 eigenvalues belong to N(A T ).
For diagonalizable A , the number of 0 eigenvalues is precisely
dim N(A T ) while the corresponding left-eigenvectors span N(A T ). The real
and imaginary parts of the left-eigenvectors remaining span R(A T ). ⋄
Proof. First we show, for a diagonalizable matrix, the number of 0
eigenvalues is precisely the dimension of its nullspace while the eigenvectors
corresponding to those 0 eigenvalues span the nullspace:
Any diagonalizable matrix A∈ R m×m must possess a complete set of
linearly independent eigenvectors. If A is full-rank (invertible), then all
m=rank(A) eigenvalues are nonzero. [287,5.1]
Suppose rank(A)< m . Then dim N(A) = m−rank(A). Thus there is
a set of m−rank(A) linearly independent vectors spanning N(A). Each
of those can be an eigenvector associated with a 0 eigenvalue because
A is diagonalizable ⇔ ∃ m linearly independent eigenvectors. [287,5.2]
Eigenvectors of a real matrix corresponding to 0 eigenvalues must be real. A.17
Thus A has at least m−rank(A) eigenvalues equal to 0.
Now suppose A has more than m−rank(A) eigenvalues equal to 0.
Then there are more than m−rank(A) linearly independent eigenvectors
associated with 0 eigenvalues, and each of those eigenvectors must be in
N(A). Thus there are more than m−rank(A) linearly independent vectors
in N(A) ; a contradiction.
Therefore diagonalizable A has rank(A) nonzero eigenvalues and exactly
m−rank(A) eigenvalues equal to 0 whose corresponding eigenvectors span
N(A).
By similar argument, the left-eigenvectors corresponding to 0 eigenvalues
span N(A T ).
Next we show when A is diagonalizable, the real and imaginary parts of
its eigenvectors (corresponding to nonzero eigenvalues) span R(A) :
A.17 Let ∗ denote complex conjugation. Suppose A=A ∗ and As i = 0. Then s i = s ∗ i ⇒
As i =As ∗ i ⇒ As ∗ i =0. Conversely, As∗ i =0 ⇒ As i=As ∗ i ⇒ s i= s ∗ i .